Mathematical background 1 Mathematical background Mathematical background 2

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Mathematical background 1 Mathematical background

Mathematical background 1 Mathematical background

Mathematical background 2 Mathematics and engineering For engineers, mathematics is a tool: – Of

Mathematical background 2 Mathematics and engineering For engineers, mathematics is a tool: – Of course, that doesn’t mean it always works. . .

Mathematical background 3 Justification However, as engineers, you will not be paid to say:

Mathematical background 3 Justification However, as engineers, you will not be paid to say: Method A is better than Method B or Algorithm A is faster than Algorithm B Such comparisons are said to be qualitative: qualitative, a. Relating to, connected or concerned with, quality or qualities. Now usually in implied or expressed opposition to quantitative.

Mathematical background 4 Justification Qualitative statements cannot guide engineering design decisions: – Algorithm A

Mathematical background 4 Justification Qualitative statements cannot guide engineering design decisions: – Algorithm A could be better than Algorithm B, but Algorithm A would require three person weeks to implement, test, and integrate while Algorithm B has already been implemented and has been used for the past year – There are circumstances where it may beneficial to use Algorithm A, but not based on the word better

Mathematical background 5 Justification Thus, we will look at a quantitative means of describing

Mathematical background 5 Justification Thus, we will look at a quantitative means of describing data structures and algorithms: quantitative, a. Relating to, concerned with, quantity or its measurement; ascertaining or expressing quantity. This will be based on mathematics, and therefore we will look at a number of properties which will be used again and again throughout this course

Mathematical background 6 Floor and ceiling functions The floor function maps any real number

Mathematical background 6 Floor and ceiling functions The floor function maps any real number x onto the greatest integer less than or equal to x: – Consider it rounding towards negative infinity The ceiling function maps x onto the least integer greater than or equal to x: – Consider it rounding towards positive infinity

Mathematical background 7 Logarithms We will begin with a review of logarithms: If n

Mathematical background 7 Logarithms We will begin with a review of logarithms: If n = em, we define m = ln( n ) It is always true that eln(n) = n; however, ln(en) = n requires that n is real

Mathematical background 8 Logarithms Exponentials grow faster than any non-constant polynomial for any d

Mathematical background 8 Logarithms Exponentials grow faster than any non-constant polynomial for any d > 0 Thus, their inverses—logarithms—grow slower than any polynomial

Mathematical background 9 Logarithms Example: is strictly greater than ln(n)

Mathematical background 9 Logarithms Example: is strictly greater than ln(n)

Mathematical background 10 Logarithms grows slower but only up to n = 93 ln(n)

Mathematical background 10 Logarithms grows slower but only up to n = 93 ln(n) (93. 354, 4. 536)

Mathematical background 11 Logarithms You can view this with any polynomial ln(n) (5503. 66,

Mathematical background 11 Logarithms You can view this with any polynomial ln(n) (5503. 66, 8. 61)

Mathematical background 12 Logarithms We have compared logarithms and polynomials – How about log

Mathematical background 12 Logarithms We have compared logarithms and polynomials – How about log 2(n) versus ln(n) versus log 10(n) You have seen the formula Because, All logarithms are scalar multiples of each others Constant

Mathematical background 13 Logarithms A plot of log 2(n) = lg(n), ln(n), and log

Mathematical background 13 Logarithms A plot of log 2(n) = lg(n), ln(n), and log 10(n) lg(n) ln(n) log 10(n)

Mathematical background 14 Logarithms You should also, as electrical or computer engineers be aware

Mathematical background 14 Logarithms You should also, as electrical or computer engineers be aware of the relationship: lg(210) = lg(1024) = 10 lg(220) = lg(1 048 576) = 20 and consequently: lg(103) = lg(1000) ≈ 10 kilo lg(106) = lg(1 000) ≈ 20 mega lg(109) ≈ 30 giga lg(1012) ≈ 40 tera

Mathematical background 15 Arithmetic series Next we will look various series Each term in

Mathematical background 15 Arithmetic series Next we will look various series Each term in an arithmetic series is increased by a constant value (usually 1) :

Mathematical background 16 Arithmetic series Proof 1: write out the series twice and add

Mathematical background 16 Arithmetic series Proof 1: write out the series twice and add each column 1 + 2 + 3 +. . . + n – 2 + n – 1 + n – 2 +. . . + 3 + 2 + 1 (n + 1) +. . . + (n + 1) = n (n + 1) Since we added the series twice, we must divide the result by 2

Mathematical background 17 Arithmetic series Proof 2 (by induction): The statement is true for

Mathematical background 17 Arithmetic series Proof 2 (by induction): The statement is true for n = 0: Assume that the statement is true for an arbitrary n:

Mathematical background 18 Arithmetic series Using the assumption that for n, we must show

Mathematical background 18 Arithmetic series Using the assumption that for n, we must show that

Mathematical background 19 Arithmetic series Then, for n + 1, we have: By assumption,

Mathematical background 19 Arithmetic series Then, for n + 1, we have: By assumption, the second sum is known:

Mathematical background 20 Arithmetic series The statement is true for n = 0 and

Mathematical background 20 Arithmetic series The statement is true for n = 0 and the truth of the statement for n implies the truth of the statement for n + 1. Therefore, by the process of mathematical induction, the statement is true for all values of n ≥ 0. Reference: Mr. Oprendick

Mathematical background 21 Other polynomial series We could repeat this process, after all: however,

Mathematical background 21 Other polynomial series We could repeat this process, after all: however, it is easier to see the pattern:

Mathematical background 22 Other polynomial series We can generalize this formula Demonstrating with d

Mathematical background 22 Other polynomial series We can generalize this formula Demonstrating with d = 3 and d = 4:

Mathematical background 23 Other polynomial series The justification for the approximation is that we

Mathematical background 23 Other polynomial series The justification for the approximation is that we are approximating the sum with an integral: n 2

Mathematical background 24 Other polynomial series The ratio between the error and the actual

Mathematical background 24 Other polynomial series The ratio between the error and the actual value goes to zero: – In the limit, as n → ∞, the ratio between the sum and the approximation goes to 1 – The relative error of the approximation goes to 0

Mathematical background 25 Geometric series The next series we will look at is the

Mathematical background 25 Geometric series The next series we will look at is the geometric series with common ratio r: and if |r| < 1 then it is also true that

Mathematical background 26 Geometric series Elegant proof: multiply by Telescoping series: all but the

Mathematical background 26 Geometric series Elegant proof: multiply by Telescoping series: all but the first and last terms cancel Ref: Bret D. Whissel, A Derivation of Amortization

Mathematical background 27 Geometric series Note that we can use a change-of-index with summations

Mathematical background 27 Geometric series Note that we can use a change-of-index with summations like we do with integration: Letting j = i – 1:

Mathematical background 28 Recurrence relations Finally, we will review recurrence relations: – Sequences may

Mathematical background 28 Recurrence relations Finally, we will review recurrence relations: – Sequences may be defined explicitly: xn = 1/n 1, 1/2, 1/3, 1/4, . . . Here n is xn independent. – Consider a fibonacci series where F 1=1, F 2=2 and F=F+F Then the series is 1, 2, 3, 5, 8, 13, 21, … – A recurrence relationship is a means of defining a sequence based on previous values in the sequence – Such definitions of sequences are said to be recursive

Mathematical background 29 Recurrence relations Define an initial value: e. g. , x 1

Mathematical background 29 Recurrence relations Define an initial value: e. g. , x 1 = 1 Defining xn in terms of previous values: – For example, xn = x n – 1 + 2 xn = 2 xn – 1 + n xn = xn – 1 + xn – 2

Mathematical background 30 Recurrence relations Given the two recurrence relations xn = x n

Mathematical background 30 Recurrence relations Given the two recurrence relations xn = x n – 1 + 2 xn = 2 xn – 1 + n and the initial condition x 1 = 1 we would like to find explicit formulae for the sequences In this case, we have xn = 2 n – 1 respectively xn = 2 n + 1 – n – 2

Mathematical background 31 Recurrence relations We will use a functional form of recurrence relations:

Mathematical background 31 Recurrence relations We will use a functional form of recurrence relations: Calculus CSE x 1 = 1. . . f(1) = 1. . . . . xn = xn – 1 + 2. . f(n) = f(n – 1) + 2. . . xn = 2 xn – 1 + n f(n) = 2 f(n – 1) + n

Mathematical background 32 Recurrence relations The previous examples using the functional notation are: f(n)

Mathematical background 32 Recurrence relations The previous examples using the functional notation are: f(n) = f(n – 1) + 2 g(n) = 2 g(n – 1) + n With the initial conditions f(1) = g(1) = 1, the solutions are: f(n) = 2 n – 1 g(n) = 2 n + 1 – n – 2

Mathematical background 33 Recurrence relations In some cases, given the recurrence relation, we can

Mathematical background 33 Recurrence relations In some cases, given the recurrence relation, we can find the explicit formula: – Consider the Fibonacci sequence: f(n) = f(n – 1) + f(n – 2) f(0) = f(1) = 1 that has the solution where f is the golden ratio:

Mathematical background 34 Weighted averages Given n objects x 1, x 2, x 3,

Mathematical background 34 Weighted averages Given n objects x 1, x 2, x 3, . . . , xn, the average is Given a sequence of coefficients c 1 , c 2 , c 3 , … , cn where then we refer to as a weighted average For an average,

Mathematical background 35 Weighted averages Examples: – Simpson’s method approximates an integral by sampling

Mathematical background 35 Weighted averages Examples: – Simpson’s method approximates an integral by sampling the function at three points: f(a), f(b), f(c) – The average value of the function is approximated by – It can be shown that is a significant better approximation than

Mathematical background 36 Weighted averages Examples: – Using the weighted average: – Using a

Mathematical background 36 Weighted averages Examples: – Using the weighted average: – Using a simple average:

Mathematical background 37 Combinations Given n distinct items, in how many ways can you

Mathematical background 37 Combinations Given n distinct items, in how many ways can you choose k of these? – I. e. , “In how many ways can you combine k items from n? ” – For example, given the set {1, 2, 3, 4, 5}, I can choose three of these in any of the following ways: {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}, The number of ways such items can be chosen is written where is read as “n choose k”s There is a recursive definition:

Mathematical background 38 Combinations The most common question we will ask in this vein:

Mathematical background 38 Combinations The most common question we will ask in this vein: – Given n items, in how many ways can we choose two of them? – In this case, the formula simplifies to: For example, given {0, 1, 2, 3, 4, 5, 6}, we have the following 21 pairs: {0, 1}, {0, 2}, {0, 3}, {0, 4}, {0, 5}, {0, 6}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}, {4, 5}, {4, 6}, {5, 6}

Mathematical background 39 Combinations You have also seen this in expanding polynomials: For example,

Mathematical background 39 Combinations You have also seen this in expanding polynomials: For example,

Mathematical background 40 Combinations These are also the coefficients of Pascal’s triangle:

Mathematical background 40 Combinations These are also the coefficients of Pascal’s triangle:

Mathematical background 41 Absolute Value or Modulus of a |a|= non-negative value of a

Mathematical background 41 Absolute Value or Modulus of a |a|= non-negative value of a Example: |5|=5 |0|=0 |-5|=5 |-5. 5|=5. 5

Mathematical background 42 Modulo the result of the modulo operation is the remainder of

Mathematical background 42 Modulo the result of the modulo operation is the remainder of the Euclidean division. In nearly all computing systems, the quotient q and the remainder r of a divided by n satisfy q is an integer number. a=nq+r |r|<|n| This is then written as r=(a mod n)

Mathematical background 43 Modulo (-a mod n)=n-(a mod n) Example: 16 mod 3=1, -16

Mathematical background 43 Modulo (-a mod n)=n-(a mod n) Example: 16 mod 3=1, -16 mod 3=2

Mathematical background 44 End of the Chapter

Mathematical background 44 End of the Chapter